1) \(\sqrt[]{3x+7}-5< 0\)

\(\Leftrightarrow\sqrt[]{3x+7}< 5\)

\(\Leftrightarrow3x+7\ge0\cap3x+7< 25\)

\(\Leftrightarrow x\ge-\dfrac{7}{3}\cap x< 6\)

\(\Leftrightarrow-\dfrac{7}{3}\le x< 6\)

1.

ĐKXĐ: \(x< 5\)

\(\Leftrightarrow\sqrt{\dfrac{42}{5-x}}-3+\sqrt{\dfrac{60}{7-x}}-3=0\)

\(\Leftrightarrow\dfrac{\dfrac{42}{5-x}-9}{\sqrt{\dfrac{42}{5-x}}+3}+\dfrac{\dfrac{60}{7-x}-9}{\sqrt{\dfrac{60}{7-x}}+3}=0\)

\(\Leftrightarrow\dfrac{9x-3}{\left(5-x\right)\left(\sqrt{\dfrac{42}{5-x}}+3\right)}+\dfrac{9x-3}{\left(7-x\right)\left(\sqrt{\dfrac{60}{7-x}}+3\right)}=0\)

\(\Leftrightarrow\left(9x-3\right)\left(\dfrac{1}{\left(5-x\right)\left(\sqrt{\dfrac{42}{5-x}}+3\right)}+\dfrac{1}{\left(7-x\right)\left(\sqrt{\dfrac{60}{7-x}}+3\right)}\right)=0\)

\(\Leftrightarrow x=\dfrac{1}{3}\)

b.

ĐKXĐ: \(x\ge2\)

\(\sqrt{\left(x-2\right)\left(x-1\right)}+\sqrt{x+3}=\sqrt{x-2}+\sqrt{\left(x-1\right)\left(x+3\right)}\)

\(\Leftrightarrow\sqrt{\left(x-2\right)\left(x-1\right)}-\sqrt{x-2}+\sqrt{x+3}-\sqrt{\left(x-1\right)\left(x+3\right)}=0\)

\(\Leftrightarrow\sqrt{x-2}\left(\sqrt{x-1}-1\right)-\sqrt{x+3}\left(\sqrt{x-1}-1\right)=0\)

\(\Leftrightarrow\left(\sqrt{x-1}-1\right)\left(\sqrt{x-2}-\sqrt{x+3}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-1}-1=0\\\sqrt{x-2}-\sqrt{x+3}=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=1\\x-2=x+3\left(vn\right)\end{matrix}\right.\)

\(\Rightarrow x=2\)

a: ĐKXĐ: -2<=x<=2

Đặt \(a=\sqrt{2+x};b=\sqrt{2-x}\)

Phương trình sẽ trở thành:

ab+6=2a+3b

=>ab-2a-3b+6=0

=>a(b-2)-3(b-2)=0

=>(b-2)(a-3)=0

=>b=2 hoặc a=3

=>2+x=9 hoặc 2-x=4

=>x=-7(loại) hoặc x=-2(nhận)

b: ĐKXĐ: x<=2

\(\left(\sqrt{2-x}+1\right)^2=3x+1\)

=>\(2-x+1+2\cdot\sqrt{2-x}=3x+1\)

=>\(-x+3+2\cdot\sqrt{2-x}-3x-1=0\)

=>\(2\cdot\sqrt{2-x}-4x+2=0\)

=>\(\sqrt{2-x}-2x+1=0\)

=>\(\sqrt{2-x}=2x-1\)

=>\(\begin{cases}2x-1\ge0\\ \left(2x-1\right)=\left(2-x\right)^2\end{cases}\Rightarrow\begin{cases}x\ge\frac12\\ x^2-4x+4=2x-1\end{cases}\)

=>\(\begin{cases}x\ge\frac12;x\le2\\ x^2-6x+5=0\end{cases}\)

=>(x-1)(x-5)=0 và 1/2<=x<=2

=>x=1